Find the Area outside the circle $r=2$ and inside the lemniscate $r^2=8 \cos(2\theta)$ where $\theta \in [0,2\pi]$
I try to graph using Mathematica and got this graph Graph of the lemniscate and circle with radius 2
First, if we substract Lemniscate Area - Circle Area it wont work because theres so many remainder of the area. $Lemniscate Area< Circle Area$ (I have check this and its $8<4\pi$)
Secondly, if we want to find the area of the lemniscate outside the circle. I think we need to find the area of the lemniscate inside the circle. But, in my opinion the only way is to find its function of polar that satisfies the circle between $[\frac{-\pi}{6},\frac{\pi}{6}]$ and $[\frac{5\pi}{6},\frac{7\pi}{6}]$ and also satisfies the lemniscate, but i dont know how to find that equation.
Is there a simple way to do this or is there anything wrong in my argument?
Looks like you are on the right track...
You have the points where the circle intersects the lemniscate
Integrate in polar coordinates over the relevant intervals.
$A = \int_{-\frac {\pi}{6}}^\frac {\pi}{6} \frac 12 (8\cos 2\theta - 4)\ d\theta + \int_{\frac {5\pi}{6}}^\frac {7\pi}{6} \frac 12 (8\cos 2\theta - 4)\ d\theta$